Approximating $\int_1^{10}x^x\mathrm dx$ to within 5% relative error

The answer can be found in the paper : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function

An asymptotic expansion of the so called Sophomores Dream function $$\text{Sphd}(\alpha;x)=\int_0^{x}t^{\alpha t} dt$$ is given in section 6 , pp.6-7.

To reach the specified accuracy, it is not necessary to use many terms of the series. Only the first term is sufficient. In fact, this is the equivalent for large $x$ given page 9, Eq.(9:2) : $$\text{Sphd}(\alpha;x)\sim \frac{x^{\alpha x}}{\alpha(1+\ln(x))}$$

In the present case, with $\alpha=1$ : $$\int_1^{10}x^x dx=\text{Sphd}(1;10)-\text{Sphd}(1;1)\simeq \frac{10^{10}}{1+\ln(10)}\simeq 3.027931\times 10^9$$

From Eq.(8:1)$\quad \text{Sphd}(1;1)\simeq 0.783430\quad$ is negligible.

One can compare to the result of numerical calculus : $\quad \int_1^{10}x^x dx\simeq 3.057489\times 10^9\quad$ The above approximate leads to a relative error lower than 1%.

Consider $$\int{x^xdx} = \int{e^{x\log (x)}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\log^k (x)}{k!}}\,dx=\sum_{k=0}^{\infty}\frac 1{k!}\int {x^k\log^k (x)}\,dx$$Now $$\int {x^k\log^k (x)}\,dx=-\log ^{k+1}(x) (-(k+1) \log (x))^{-k-1}\,\Gamma (k+1,-(k+1) \log (x))$$ where appears the incomplete gamma function.

Using the bounds, the table below reproduces the results summing from $k=0$ to $k=n$ $$\left( \begin{array}{cc} n &\sum_{k=1}^{n} \\ 10 & 1.34986\times 10^7 \\ 15 & 2.49883\times 10^8 \\ 20 & 1.186883\times 10^9 \\ 25 & 2.35791\times 10^9 \\ 30 & 2.92299\times 10^9 \\ 35 & 3.04399\times 10^9 \\ 40 & 3.05675\times 10^9 \\ 45 & 3.05747\times 10^9 \\ 50 & 3.05749\times 10^9 \end{array} \right)$$ which is the solution for six significant figures.